Your compound is

.
Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.
Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4
Answer:
3.7 mol Al2O3 x 4 mol Al = 7.4 mol Al 2 mol Al2O3
Explanation:
Basis: 100 mL solution
From the given density, we calculate for the mass of the solution.
density = mass / volume
mass = density x volume
mass = (1.83 g/mL) x (100 mL) = 183 grams
Then, we calculate for the mass H2SO4 given the percentage.
mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams
Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles
Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M
Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m
Empirical formula of a compound gives the proportions of the elements in that compound but it does not define the actual arrangement and number of atoms.
Let the empirical formula of compound be
.
The ratio of number of moles of C, H and O can be calculated as follows:

Simplifying the ratio,

Thus, the value of x, y and z will be 1, 2 and 1 respectively.
Therefore, the empirical formula will be
.
Answer:
The half-life varies depending on the isotope.
Half-lives range from fractions of a second to billions of years.
The half-life of a particular isotope is constant.